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Proof step 60 from proof square-neq-any-pos-02 (previous step)

0 < j1 * abs(i0), prime(j1), j1 ≤ j1 * abs(i0), 1 < j1 * j1 * i0 * i0, 0 < j1 * j1 * i0 * i0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * j1 * i0 * i0) → 0 < i1 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i1 * j0 * j0 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i. j1 * abs(i0) = abs(i)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: j1 * i0 abort: F discard: T computed: F

Additional information

Used simplifier rules: prime-abs: ⊦ prime(j) → abs(j) = j;%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j1 * i0